Mistake Master
Mistake Master
Push on a surface and you exert pressure: the perpendicular force per unit area, $P = F_\perp / A$. In a still fluid, pressure rises with depth as $P = P_0 + \rho g h$. The trap: at a given depth, pressure depends only on $h$ and $\rho$. Not the container shape. Not how much fluid sits above.
Pressure is force per unit area. The "force" is just the perpendicular part (the push straight into the surface), and the "area" is what it's spread over:
$$P \;=\; \dfrac{F_\perp}{A}$$The unit is the pascal: $1 \text{ Pa} = 1 \text{ N/m}^2$. Pressure is a scalar. It has a size but no direction, even though the force pushing on a surface does.
This is why a 60 kg person in stiletto heels presses harder on a floor than a 200 kg piano on four wide casters. The piano weighs more, sure, but it spreads that weight across way more area. The heels concentrate a smaller weight onto tiny tips, and tiny area is what bites.
One more thing worth a name: push on a fluid that's trapped in a container, and the pressure jumps by the same amount everywhere inside (Pascal's principle). This is how hydraulic systems work; the AP exam won't ask you to crunch numbers on them.
In a still fluid that's open to the air on top, the pressure at depth $h$ below the surface is
$$P \;=\; P_0 \;+\; \rho g h$$Here $P_0$ is the pressure at the surface (usually atmospheric, $P_0 = P_\text{atm}$), $\rho$ is the fluid density, $g$ is gravitational field strength, and $h$ is how far down you've gone.
Where the formula comes from. Picture a vertical column of fluid, cross-section $A$, running from the surface down to depth $h$. Its weight is $W = (\rho A h) g$. That weight presses on area $A$ at the bottom, adding $W/A = \rho g h$ of pressure on top of whatever was at the surface. The column directly above the point is what matters; the rest of the container does not.
What doesn't matter. Container shape. Total fluid volume. Footprint area. A tall narrow tube and a wide reservoir filled to the same depth read the same pressure at the bottom. This is called the hydrostatic paradox; it feels wrong, but the formula is dead clear.
A swimmer is 5 m below the surface of a freshwater lake ($\rho = 1000 \text{ kg/m}^3$). Use $g = 10 \text{ N/kg}$. Find the gauge pressure of the water on her.
Setup. Gauge means relative to atmospheric, so $P_\text{gauge} = \rho g h$.
Plug in. $P_\text{gauge} = (1000)(10)(5) = 50{,}000 \text{ Pa} = 50 \text{ kPa}$.
Read it. That's 50 kPa pushing on every square meter of her skin, in every direction. About half an atmosphere of extra squeeze, just five meters down.
$P_\text{gauge} = \rho g h$ depends on $\rho$ and $h$. Not on shape, not on volume, not on how much fluid sits to either side.
A pressure number is always read against some zero. Two conventions show up:
Absolute pressure measures from a true vacuum. The air around you sits at about $P_\text{atm} = 1.0 \times 10^5 \text{ Pa}$ absolute. Gauge pressure measures from atmospheric. The relation:
$$P_\text{gauge} \;=\; P_\text{abs} \;-\; P_\text{atm}$$A tire gauge that reads $30 \text{ psi}$ is reporting gauge pressure. The absolute pressure inside is the gauge reading plus atmospheric, about $30 + 14.7 = 44.7 \text{ psi}$. When the tire is flat and the gauge reads zero, the inside is still at one atmosphere; the gauge has just hit its zero point.
For a fluid open to the air at the top with $P_0 = P_\text{atm}$, the formula $P = P_0 + \rho g h$ gives absolute pressure. Drop the $P_\text{atm}$ and you get the gauge form, $P_\text{gauge} = \rho g h$. Most depth-pressure problems use gauge by default; sealed containers and barometers use absolute. Read the question first and decide which one it wants.
Gauge tells you how far above atmospheric you are; absolute is the total. They differ by exactly $P_\text{atm}$. Always check which one the problem wants.
It feels right that a wide reservoir, holding way more water, should hit harder at the bottom than a narrow tube of the same depth. The reservoir does hold more weight at the bottom, that part is true. But pressure is weight per area, and a wider bottom spreads that extra weight over more area. The ratio comes out the same.
Fix. $P = \rho g h$. Only depth and fluid density. The shape and the amount drop out of the formula, every time.
Force and pressure are not the same thing. A 10 N force packed onto a square millimeter is enormous pressure; the same 10 N spread over a square meter is barely anything. The force alone can't tell you which one you're in; you need the area too.
Fix. $P = F_\perp / A$. Always carry the area through. A small force on a tiny area can be huge pressure; a large force on a large area can be modest pressure.
The gauge reads pressure above atmospheric, not total pressure. When the tire is flat the gauge reads zero, but the inside is still at about one atmosphere. Confusing gauge with absolute is a one-atmosphere error every time it happens.
Fix. $P_\text{abs} = P_\text{atm} + P_\text{gauge}$. Decide up front which one the problem wants, then convert if needed.
Ten scenarios mixing focus-and-choice, symbolic, and numeric prompts. Each one targets a specific way pressure goes sideways. Pick the right option and you'll see the trap baked into the others. Your progress saves in the browser as you go.